03. Diffraction Integrals (Fresnel + Fraunhofer propagation, Point Spread Function, Fourier optics)

[@joem8251]

I am blown away by how quickly, clearly, and concisely you present this material!

[@shane3379]

I am only 5 minutes into this video and have already learned more than in my entire advanced optics module.... brilliant video, this is the future of learning

[@worldpulse365]

I'm still learning a lot of fourier optics concepts...going over LARGE, demoralizing textbooks😑 and you just managed to compress so much into one video. Incredible!

[@ovieigherebuo9317]

Thanks a whole semesters lecture in one coherent video

[@martinmillischer6083]

Last year I took a class at photonics west on Fourier Optics. It was 8h long and $600. In 25 min I learn more with your video. Thank you so much for your work.

[@jacobvandijk6525]

@ 0:55 Here, in H's Principle, the source of light is called a field too. One could call this starting field U(0,0). @ 1:40 If you also put in U(0,0), then you have an expression for "the propagator" from QFT. Here, U(0,0) = 1.

[@iosianchad2180]

Good material for reviewing wave optics after dropping it for 20 years.

[@Kolibril]

Thank you, this is really condensed and purified content

[@arturocatalanogonzaga7874]

Hi! Thanks a lot for the extremely helpful videos! Could you please clarify how and why, at minute 21:16, you went from having an Object in the (xo,yo) object plane to the (x/M,y/M) plane? why is it the lens plane that gets demagnified? shouldn't it be (xo/M,yo/M)?

[@SanderKonijnenberg]

My apologies for the confusion: in this slide, (x,y) denotes the image plane coordinates (not the lens plane). If we ignore the effects of the PSF (i.e. assume that it's a delta peak), then the image I(x,y) should be a magnified copy of |O(x_o,y_o)|^2. That is, I(x,y)=|O(x/M,y/M)|^2, so that the object field at object coordinates (x_o,y_o) is mapped to image coordinates (x,y)=(M*x_o,M*y_o).

[@arturocatalanogonzaga7874]

@@SanderKonijnenberg thank you so much for the reply!

[@grandaurore]

3:58 why in approximation on complex exponential, we should change the sign?

[@SanderKonijnenberg]

Because exp(ik * lambda/2)=exp(i*pi)=-1 (recall k=2pi/lambda)

[@grandaurore]

thanks, very clear, the minus come from exp(i*pi). I was misleading to expansion generating the minus sign.

[@mementomori6734]

18:06 from angular spectrum method we have that the resolution is limited to lambda/2. But if we have a lens with a NA>1 we can have a resolution

[@SanderKonijnenberg]

Generally, NA is defined NA=n*sin(theta), where n is the refractive index of the immersion medium. So NA>1 is typically achieved by immersing the sample in a medium with high refractive index, e.g. water or immersion oil. If the refractive index of a medium is n, then the wavelength of light in that medium is lambda/n (where lambda is the wavelength in vacuum). By decreasing the wavelength we increase the resolution. Therefore, the resolution is still larger than lambda/2 if we define lambda to be the wavelength in the immersion medium, but it can be smaller than lambda/2 if we define lambda to be the wavelength in vacuum.

[@Lovearbre]

Great Video! Thank you!

[@rajatsaxena1703]

Great video!

[@한두혁]

What does global phase factor in 10:36 mean? Thank you for this wonderful video!

[@SanderKonijnenberg]

Thanks for your comment. 'Global phase factor' means in this case a phase factor that does not depend on (x,y), and therefore does not affect the way that the field propagates.

[@daguaishouxd]

Just a minor question, at 15:30, I thought that in F.T. shift theorem the shift in F.T. has the same sign as the exponent of the phase shift, i.e. F{f(x)e^(-2i pi ax)}(x',y') = F{f(x)}(x'-a, y'-a). I guess the later equations will still hold if we define M as |z_i|/|z_0| without the negative sign... Any thoughts?

[@SanderKonijnenberg]

If the Fourier transform is defined as F{f(x)} (x') = int f(x)e^(-2 pi i x x') dx then F{f(x) e^(- 2 pi i ax)} (x') = int f(x)e^(-2 pi i x (x'+a)) dx = F{f(x)} (x'+a). Perhaps what you're thinking of is the inverse Fourier transform? If you multiply F{f(x)} (x') with e^(-2 pi i a x'), then inverse Fourier transforming get you f(x-a). Physically, we also know that a real image generated by a single lens is inverted, so one way or another, M must turn out to be negative.

[@daguaishouxd]

@@SanderKonijnenberg Yes you are definitely right. It's easier to think of (x'+a) as a new dummy variable then it's quite obvious. I did confused it with an e^(-2 pi i a x') shift in x' domain which will lead to f(x-a) in x domain. Thanks for clarifying!

[@mightbin]

at 2:23 , i think the complex form of the wave function i thinkl is : z/r2 * ei(kr−ωt), why do you say it is z/r2 * e ikr, and ommit the −ωt part ? Thanks a ot.

[@SanderKonijnenberg]

Yes, technically the field has a time-dependence, but the advantage of complex notation is that it allows you to omit the time-dependent complex exponential e^{-i\omega t}, see my video kzbin.info/www/bejne/aWKTaGWgi8t8i6c at 7:39 - 9:33 . For a monochromatic field, we typically don't care about how each point oscillates in time, but only about the phase difference between fields at different points in space.

[@mightbin]

really helpful. Thank you so much.@@SanderKonijnenberg

[@LixiaCao-l5l]

Can you share the MATLAB code?

[@SanderKonijnenberg]

The Google drive link in the video description also contains a zip-file with the Matlab codes. Hopefully those include what you're looking for.

[@nebulae_wanderer]

My 5am savior

[@thetatheta9926]

Whats the difference between Rayleigh-Sommerfeld integral and Huygen-Fresnel integral?

[@SanderKonijnenberg]

That's a good question, I never thought about it before. Section 3.7 of Joseph Goodman's 'Introduction to Fourier Optics' (can be found online) says: "The Huygens-Fresnel principle, as predicted by the first Rayleigh-Sommerfeld solution (see Eq. (3-40)), can be expressed mathematically as follows: [...]". It appears that the Huygens-Fresnel integral is a heuristically derived principle, whereas the Rayleigh-Sommerfeld solution is more rigorously derived from the wave equation, and from which the Huygens-Fresnel principle follows. But ultimately, it appears they describe the same thing.

[@thetatheta9926]

@@SanderKonijnenberg Thanks for the reply. I'm actually trying to accurately determine how the field of a Gaussian beam is transformed after impinging on a reflective metasurface (i.e. is it still a Gaussian beam?). I applied The Huygen-Fresnel diffraction integral mainly because it appeared to be a simpler version of the RS integral. I guess I have to checkout Goodman.

[@adurgh]

any reference textbooks on this that can be recommended?

[@SanderKonijnenberg]

I'd go with J. Goodman's 'Introduction to Fourier Optics'

[@esmirhodzic981]

very nice video

[@ErichAmMeer]

why are that many books so long and boring without clear explanation，disappointing until I found this